Let $R$ be the region enclosed by the line $x=-4$, the line $y=1$, the line $y=4$, and the curve $y=e^x$. $y$ $x$ ${y=e^x}$ $ -4$ $ R$ $ 1$ $ 4$ A solid is generated by rotating $R$ about the line $x=-4$. Which one of the definite integrals gives the volume of the solid? Choose 1 answer: Choose 1 answer: (Choice A) A $\pi \int_1^4 \left[\text{ln}(y)+4\right]^2dy$ (Choice B) B $\pi \int_1^4 \left[\text{ln}(y)-4\right]^2dy$ (Choice C) C $\pi \int_1^4 \left[e^y-4\right]^2dy$ (Choice D) D $\pi \int_1^4 \left[e^y+4\right]^2dy$
Answer: Let's imagine the solid is made out of many thin slices. $y$ $x$ ${y=e^x}$ $ -4$ Notice the slices are horizontal, because we are rotating $R$ about a vertical axis. Each slice is a cylinder. Let the thickness of each slice be $dy$ and let the radius of the base, as a function of $y$, be $r(y)$. Then, the volume of each slice is $\pi [r(y)]^2\,dy$, and we can sum the volumes of infinitely many such slices with an infinitely small thickness using a definite integral: $\int_a^b \pi [r(y)]^2\,dy$ This is called the disc method. What we now need is to figure out the expression of $r(y)$ and the interval of integration. Let's consider one such slice. $y$ $x$ ${y=e^x}$ $ -4$ $ 1$ $ 4$ $r$ The radius is equal to the distance between the curve ${y=e^x}$ and the line ${x=-4}$. To find it, we need to solve $y=e^x$ for $x$ : ${x=\text{ln}(y)}$ So, for any $y$ -value, this is the equation for $r(y)$ : $\begin{aligned} r(y)}&={\text{ln}(y)}-({-4}) \\\\ &=\text{ln}(y)+4} \end{aligned}$ Now we can find an expression for the area of the cylinder's base: $\begin{aligned} &\phantom{=}\pi [r(y)}]^2 \\\\ &=\pi\left[\text{ln}(y)+4}\right]^2 \end{aligned}$ The bottom endpoint of $R$ is at $y=1$ and the top endpoint is at $y=4$. So the interval of integration is $[1,4]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int_1^4 \pi\left[\text{ln}(y)+4\right]^2dy \\\\ &=\pi \int_1^4 \left[\text{ln}(y)+4\right]^2dy \end{aligned}$